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chwala
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- Homework Statement
- Prove that ##\csc ∅ - \cot ∅ = \tan \dfrac {∅}{2}##. Hence, deduce that ##\tan \dfrac {π}{8} = \sqrt{2} -1## and find ##\tan \dfrac {3π}{8}## in terms of ##a+ b\sqrt{2}##
- Relevant Equations
- Trigonometry.
Refreshing on trig. today...a good day it is...ok find the text problem here; With maths i realize one has to keep on refreshing at all times... my target is to solve 5 questions from a collection of 10 textbooks i.e 50 questions on a day-day basis...motivation from late Erdos ...( pure math, applied, stats, mechanics} each and every day starting today...i will only post the interesting questions...
Ok my approach; let ##\tan \dfrac{∅}{2}=t##
##\dfrac{1}{\sin ∅} - \dfrac{\cos∅ }{\sin ∅}=t##
##\dfrac{1-\cos ∅}{\sin ∅}=t##
##\dfrac{1+t^2-(1-t^2)}{1+t^2} ⋅\dfrac{1+t^2}{2t}=t## thus true.
for second part;
Let ##\dfrac{∅}{2}=\dfrac{π}{8}##
it follows that ##∅ = \dfrac{π}{4}##
therefore; ##\dfrac{1}{\sin \dfrac{1}{4} π} - \dfrac{\cos\dfrac{1}{4}π }{\sin \dfrac{1}{4}π} ##=## \sqrt{2} ## ##-\dfrac {1}{\sqrt{2}}## ⋅## \sqrt{2}##= ##\sqrt{2}-1##
Lastly;
##\tan \dfrac{3π}{8}= \tan \dfrac{π}{4} +\tan \dfrac{π}{8}##
=## \dfrac{\tan \dfrac{π}{4} +\tan \dfrac{π}{8}}{1-\tan \dfrac{π}{4} ⋅\tan \dfrac{π}{8}}##
=##\dfrac{1+\sqrt{2}-1}{1-\sqrt{2}+1} ##
= ##\dfrac{\sqrt{2}}{2-\sqrt{2}}## ⋅ ##\dfrac{2+\sqrt{2}}{2+\sqrt{2}}##
=##\dfrac{2+2\sqrt{2}}{2}= 1+\sqrt{2}##
Of course i know there might be a slightly different approach...would appreciate any comments. Cheers guys!
Ok my approach; let ##\tan \dfrac{∅}{2}=t##
##\dfrac{1}{\sin ∅} - \dfrac{\cos∅ }{\sin ∅}=t##
##\dfrac{1-\cos ∅}{\sin ∅}=t##
##\dfrac{1+t^2-(1-t^2)}{1+t^2} ⋅\dfrac{1+t^2}{2t}=t## thus true.
for second part;
Let ##\dfrac{∅}{2}=\dfrac{π}{8}##
it follows that ##∅ = \dfrac{π}{4}##
therefore; ##\dfrac{1}{\sin \dfrac{1}{4} π} - \dfrac{\cos\dfrac{1}{4}π }{\sin \dfrac{1}{4}π} ##=## \sqrt{2} ## ##-\dfrac {1}{\sqrt{2}}## ⋅## \sqrt{2}##= ##\sqrt{2}-1##
Lastly;
##\tan \dfrac{3π}{8}= \tan \dfrac{π}{4} +\tan \dfrac{π}{8}##
=## \dfrac{\tan \dfrac{π}{4} +\tan \dfrac{π}{8}}{1-\tan \dfrac{π}{4} ⋅\tan \dfrac{π}{8}}##
=##\dfrac{1+\sqrt{2}-1}{1-\sqrt{2}+1} ##
= ##\dfrac{\sqrt{2}}{2-\sqrt{2}}## ⋅ ##\dfrac{2+\sqrt{2}}{2+\sqrt{2}}##
=##\dfrac{2+2\sqrt{2}}{2}= 1+\sqrt{2}##
Of course i know there might be a slightly different approach...would appreciate any comments. Cheers guys!
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